Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int s, a, b;
  scanf("%d %d %d", &s, &a, &b);
  int n;
  scanf("%d", &n);
  vector<pair<int, int>> segments(n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &segments[i].first, &segments[i].second);
  }
  vector<int> rolling(n);
  for (int i = 0; i < n; i++) {
    rolling[i] = (i - 1 >= 0) ? rolling[i - 1] : 0;
    rolling[i] += segments[i].second - segments[i].first;
  }
  int q;
  scanf("%d", &q);
  for (int _ = 0; _ < q; _++) {
    int x, y;
    scanf("%d %d", &x, &y);
    double multiplier = ((double)y - s) / y;
    double beginx = (a - x) / multiplier + x;
    double endx = (b - x) / multiplier + x;
    int l = -1;
    {
      int r = n;
      while (r - l != 1) {
        int m = (l + r) / 2;
        if (segments[m].second > beginx) {
          r = m;
        } else {
          l = m;
        }
      }
      l++;
    }
    double result = 0;
    if (l < n && segments[l].first < endx) {
      if (segments[l].first <= beginx) {
        result += segments[l].second - beginx;
      } else {
        result += segments[l].second - segments[l].first;
      }
    }
    int l2 = -1;
    {
      int r = n;
      while (r - l2 != 1) {
        int m = (l2 + r) / 2;
        if (segments[m].first < endx) {
          l2 = m;
        } else {
          r = m;
        }
      }
    }
    if (l2 >= 0 && segments[l2].second > beginx) {
      if (segments[l2].second >= endx) {
        result += endx - segments[l2].first;
      } else {
        result += segments[l2].second - segments[l2].first;
      }
    }
    if (l2 > l) {
      result += rolling[l2 - 1];
      result -= rolling[l];
    }
    if (l2 == l) result -= segments[l].second - segments[l].first;
    printf("%lf\n", result * multiplier);
  }
  return 0;
}